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For a pressure-free flowing poloidal flux surface, subtracting its angular velocity times the azimuthal momentum equation from the mechanical energy equation shows that ${ϵ}_J$ is constant. It equals poloidal kinetic energy plus $(u_{\varphi }-R\omega {)}^2/2+\Phi -R^2{\omega }^2/2$. This is the total magnetohydrodynamic Bernoulli invariant minus $\omega$ times the total magnetohydrodynamic angular-momentum invariant, not the total energy alone.

On one poloidal flux surface, the field-line angular velocity $\omega (\psi )$ is constant. In a frame rotating at that rate the relative velocity is $\mathbf{v}=\mathbf{u}-R\omega {\mathbf{e}}_{\varphi }=k\mathbf{B}/\rho$, parallel to the magnetic field. The Lorentz force is perpendicular to this motion, and the Coriolis force also does no work. Centrifugal acceleration comes from the effective potential $-R^2{\omega }^2/2$. Since entropy is constant along the surface, pressure work integrates to specific enthalpy $w$, with $dw=dp/\rho$ along the flow.

Therefore kinetic energy relative to the rotating poloidal flux surface, enthalpy and effective gravitational-plus-centrifugal potential have a constant sum:This is a corotating energy integral, not the laboratory energy alone: Maxwell stress tensor stresses can transfer energy and angular momentum in the laboratory frame. In fact the expression equals $|\mathbf{u}{|}^2/2+w+\Phi -\omega R{u}_{\varphi }$, namely the total magnetohydrodynamic Bernoulli invariant minus $\omega \ell$. Differential rotation between surfaces does not invalidate the argument, because the chosen angular velocity is fixed along the particular surface being followed.

The flux representation gives $B_R=-{\psi }_z/R$ and $B_z={\psi }_R/R$. Therefore $\nabla \cdot \mathbf{B}=0$ and $\mathbf{B}\cdot \nabla \psi =0$. With magnetohydrodynamic mass loading $k=k(\psi )$,Thus the steady continuity equation is satisfied.

The cross product in the ideal magnetohydrodynamic induction equation simplifies toHence steady induction requires $\mathbf{B}\cdot \nabla (v/R)=0$. On a regular connected poloidal flux surface, this integrates locally toThis is the field-line angular velocity of an axisymmetric wind. It need not equal the gas angular velocity when $B_{\varphi }\ne 0$, because $u_{\varphi }=R\omega +k{B}_{\varphi }/\rho$. The regular-surface qualification matters: if $\psi$ is constant and the field is purely toroidal, induction alone cannot turn an arbitrary $v(R,z)/R$ into a function of $\psi$.

Put $\mathbf{F}=(\nabla \times \mathbf{B})\times \mathbf{B}$ and $E=|\mathbf{u}{|}^2/2+\Phi$. Dot the pressure-free momentum equation with $\mathbf{u}$. Since $\mathbf{B}\cdot \mathbf{F}=0$, axisymmetry givesThe azimuthal momentum equation in cylindrical coordinates similarly givesSubtracting $\omega$ times the second relation from the first, and using $\mathbf{B}\cdot \nabla \omega =0$, proves that $E-\omega R{u}_{\varphi }$ is constant along a flowing regular poloidal flux surface. For nonzero mass loading its corotating energy invariant of an axisymmetric magnetic wind isThis is the invariant labeled $ϵ$ in the question. It is a corotating mechanical energy, distinct from the total magnetohydrodynamic Bernoulli invariant, which includes electromagnetic energy flux. In standard notation it equals total specific energy minus $\omega$ times the total specific angular-momentum invariant.

Near the footpoint, negligible $k{B}_{\varphi }/\rho$ gives $u_{\varphi }\simeq R\omega$, while the poloidal velocity is parallel to the poloidal magnetic field. The gas approximately corotates and slides along a rotating field line as a bead on a wire, with effective potentialA decreasing effective potential converts corotating potential energy into poloidal kinetic energy. This is the mechanism of magnetocentrifugal acceleration; local launching does not alone guarantee global escape.

For an attractive point mass, take $\Phi =-GM/(R^2+z^2{)}^{1/2}$ and ${\omega }^2=GM/R_0^3$. Parameterize the outward straight line by $R=R_0+s\sin \alpha$, $z=s\cos \alpha$. The first derivative of ${\Phi }_{\mathrm{eff}}$ vanishes at $s=0$ because the footpoint is in circular radial balance. The meridional Hessian there is ${\omega }^2\mathrm{diag}(-3,1)$, soThus the thirty-degree magnetocentrifugal launching criterion in its strict negative-curvature form isAbove this angle the equilibrium is unstable to an outward displacement; below it the near-footpoint potential initially rises. The source's proportionality constant must be negative to describe attraction.

The marginal straight-line launch at thirty degrees has zero quadratic curvature. For precisely the straight line and point-mass potential here, let $x=s/R_0$ and $a=\sin \alpha$. Direct expansion givesAt $\alpha =\pi /6$, this is $-7x^3/16+O(x^4)$, so the outward direction is downhill at cubic order. Consequently the strict inequality is the nondegenerate quadratic criterion; it should not be interpreted as excluding every one-sided nonlinear marginal launch at equality. An exactly resting bead remains an equilibrium until displaced.

The divergence-free poloidal field can be represented by the poloidal magnetic flux functionIn a steady axisymmetric ideal magnetohydrodynamics flow, the azimuthal component of $\nabla \times (\mathbf{u}\times \mathbf{B})=0$ makes ${\mathbf{u}}_p$ parallel to ${\mathbf{B}}_p$. WriteMass conservation and $\nabla \cdot \mathbf{B}=0$ then imply $\mathbf{B}\cdot \nabla k=0$, so the magnetohydrodynamic mass loading $k=k(\psi )$ is constant along each magnetic line.

The poloidal part of $\mathbf{u}\times \mathbf{B}$ isIts curl vanishes only if its coefficient is a flux function, giving the field-line angular velocityThe azimuthal component of momentum conservation is a divergence of matter and magnetic angular-momentum flux. Dividing its field-line constant by the mass loading yields the magnetohydrodynamic angular-momentum invariantThe conservative total-energy equation similarly gives the magnetohydrodynamic Bernoulli invariantFinally, the entropy advection equation and ${\mathbf{u}}_p\parallel {\mathbf{B}}_p$ imply $s=s(\psi )$. Thus $k,\omega ,\ell ,ϵ$, and $s$ are constant along each magnetic field line.

Along the open line, $B_p\sim F/r^2$ and the mass-loading relation with $u_p\to u_{\infty }$ givesThe solutions of part b consequently haveThe azimuthal Alfvén speed therefore approaches the nonzero constantFor an unconfined outflow it is natural to take $\Phi \to 0$ and $h\to 0$ at infinity; also $u_{\varphi }\to 0$. The magnetic term in the magnetohydrodynamic Bernoulli invariant tends toHence the asymptotic energy of a radial magnetohydrodynamic wind is